# How area of transistor increases load current in current mirror circuits?

#### antonyfrancisp

Joined Mar 14, 2016
5
In the current mirror chapter of All About Circuits, in the last part it is said that to get twice load current increase area of the transistor. How is this possible. How does emitter current in a current mirror depend upon area of the transistor?

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
IC designers avoid some resistors by replacing load resistors with current sources. A circuit like an operational amplifier built from discrete components will have a few transistors and many resistors. An integrated circuit version will have many transistors and a few resistors. In Figure below One voltage reference, Q1, drives multiple current sources: Q2, Q3, and Q4. If Q2 and Q3 are equal area transistors the load currents Iload will be equal. If we need a 2·Iload, parallel Q2 and Q3. Better yet fabricate one transistor, say Q3 with twice the area of Q2. Current I3 will then be twice I2. In other words, load current scales with transistor area.
This discussion involves creating the circuit, i.e. building the transistors, in the silicon itself. This is the task of the integrated circuit (IC) designer. Most of us will never see transistors or circuits at that level, we buy them only after they are packaged. Therefore, we do not have the ability to change the physical size of the individual transistor, but can choose a different transistor to match the specification of our individual circuit.

The paragraph above uses examples of ways to get more current than what is generated by the current mirror voltage reference. First, add a second transistor to help drive the load. In the IC designer world, this is the same as changing the physical size of a single transistor in silicon to get the same effect.

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
How does emitter current in a current mirror depend upon area of the transistor?
Think of it as: if you change the transistor area you are also changing the transistor gain.

#### antonyfrancisp

Joined Mar 14, 2016
5
Think of it as: if you change the transistor area you are also changing the transistor gain.
Thank you... now i am kind of getting it

#### dl324

Joined Mar 30, 2015
12,258
if you change the transistor area you are also changing the transistor gain.
Transistor beta is primarily a function of the doping concentrations in the base and emitter regions.

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
Thank you, I misstated what I was trying to say since I was trying to keep it simple. Changing the transistor area changes its current carrying capability.

#### antonyfrancisp

Joined Mar 14, 2016
5
Thank you, I misstated what I was trying to say since I was trying to keep it simple. Changing the transistor area changes its current carrying capability.
And gain also??

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
The larger current going through the transistor is due to a multiple loads (larger current draw) now being attached to one place instead of two as with two transistors. Changing the area allows the transistor to handle the larger load.

#### dl324

Joined Mar 30, 2015
12,258
And gain also??
Yes, but in the current mirrors you were referring to, larger area gives more current at the same junction voltage.

#### antonyfrancisp

Joined Mar 14, 2016
5
Yes, but in the current mirrors you were referring to, larger area gives more current at the same junction voltage.
The larger current going through the transistor is due to a multiple loads (larger current draw) now being attached to one place instead of two as with two transistors. Changing the area allows the transistor to handle the larger load.
Actually aren't current mirrors acting like constant current sources(for a constant junction voltage). As per I understood without changing the junction voltage, current won't increase.
Is there a condition that Rbias and Rload should be equal in order to attain this constant current?
Is that why when load increases current increasing?

#### antonyfrancisp

Joined Mar 14, 2016
5
Suppose if we attach 2 more parallel loads to the transistor which demands more current.
The current we will be getting in the loop will be (Vsupply-Vbe)/Rbias. As long as Vbe is constant, this will be constant.
So what I am getting is this current gets divided and goes into different loads as per corresponding loads. Is that right?

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
Yesterday is one of those days that I should have taken my own advise and stayed at home.

For a given Vbe, the collector current of a bjt is fundamentally tied to the area of the transistor. It has a proportional relationship.

#### Bordodynov

Joined May 20, 2015
2,744
See